Find a set A (subset of R,set of real numbers) and an element a of R

julia89
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Find a set A (subset of R,set of real numbers) and an element a of R
such that there is no bijecton from a+A(we add a to the set A)to A.

I can't find a good example. Can someone help
Are we done if we choose the empty set? (And is the empty set a subset of R?)

Thank you
 
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?? Any finite set will do. Take A= {1, 2, 3} and a= 0. Then A has 3 members while A+ a has 4. There cannot be a bijection from one to the other.

Yes, the empty set is a subset of R (it is a subset of any set). Choosing A= {} and a any real number so that A+ a= {a} will also work.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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